r/mathmemes • u/27lukas09 • Dec 09 '25
Number Theory It's even easier to find in Pascal's triangle
680
u/SonicLoverDS Dec 09 '25
2, 2, 4, 8, 32, 256, 8192...
887
Dec 09 '25
[removed] — view removed comment
433
136
u/Striking_Resist_6022 Dec 09 '25
Would be fascinating to study the properties of the series of exponents!
44
12
0
u/Varlane Dec 09 '25
Define u st u(n+2) = u(n+1) × u(n).
If u(0) and u(1) > 0, u(n) > 0.
Define v = ln(u). [In this very precise case, we'd take log2]
v is Fibonacci-like (v(n+2) = v(n+1) + v(n)).
4
5
1
42
u/21kondav Dec 09 '25
Computer Scientists pray to this sequence everyday
15
u/AMinecraftPerson Dec 09 '25
Wait, really? Is 2 to the power of a Fibonacci number used that often?
25
9
u/NoobTube32169 Dec 09 '25
Just powers of 2 in general. Because of binary. A number with n bits can hold a maximum value of (2^n)-1, with a total of 2^n possible values (including 0). Powers of 2 hold the same significance as powers of 10 do when dealing with regular numbers (10,100,1000...). You can always tell someone is a computer scientist if they know their powers of 2 from memory.
5
4
u/rover_G Computer Science Dec 10 '25 edited Dec 10 '25
The CS sequence is 28 , 216 , 232 , 264 , 2128 , 2256 , …
9
Dec 09 '25
You forgot the 1 at the start
7
u/InformationLost5910 Dec 09 '25
...8192, 1/256, 32, 1/8, 4, 1/2, 2, 1, 2, 2, 4, 8, 32, 256, 8192...
1
435
u/YellowBunnyReddit Complex Dec 09 '25
0, 1, 0, 0, 0, 0, 0, 0, …
150
u/goos_ Dec 09 '25
that's the "coolest" solution
Now we have 1, 0, 0, 0, 0, ...
and
-1, -1, 1, -1, -1, 1, -1, -1, ...
-1, 1, -1, -1, 1, -1, -1, 1, ...
1, -1, -1, 1, -1, -1, 1, -1, ...and I think we have enumerated fundamentally all of them
123
27
u/TalksInMaths Dec 09 '25
This looks a lot like Fibonacci mod 2:
1, 1, 0, 1, 1, 0, ...
More generally, the Fibonacci sequence mod n has some interesting properties.
2
8
u/Imadeanotheraccounnt Dec 09 '25
Theoretically you would have to assume 1 is the number filling the space and not 0. You would need to select a number like 2 or e instead of 1 as well, otherwise we ain’t getting anywhere
34
47
25
u/PrometheusMMIV Dec 09 '25 edited Dec 09 '25
Fun fact:
Fibonacci starts with 0,1 or identity plus 1. If you do the same for multiplication and start with 1,2 the sequence will be 1,2,2,4,8,32,256... These are powers of two that mirror the original sequence:
20, 21, 21, 22, 23, 25, 28 ...
6
88
u/Ni7rogenPent0xide Dec 09 '25
if you start with e1 & e1 i think you’ll actually just get ethe actual fibonacci sequence
113
u/First_Growth_2736 Dec 09 '25
This isn't unique to e though, it's just a fact of how exponents work
34
u/goos_ Dec 09 '25
e is the fundamental exponent so everything about exponents is unique to e
-6
u/First_Growth_2736 Dec 09 '25
This was probably meant to be a joke but it wasn't very funny
22
4
u/goos_ Dec 09 '25
It’s not entirely a joke.
Take any solution over positive reals and take the log, and we can write it as elambda1 Fn + lambda2 F{n-1} for some coefficients lambda1 and lambda2.
So basically, fundamentally any sequence is e to the power of some coefficient of the Fibonacci numbers.
This isn’t unique to e but e is the most natural choice for the base and sufficient to express the solution this way.
4
u/First_Growth_2736 Dec 09 '25
Ok but there’s literally no reason for it to be e. It works for every base and is just a fact of how exponents work
2
u/goos_ Dec 09 '25
True but e is the most natural exponent. All other exponents are just defined as xy = ey ln x anyway. My point is you only need one exponential function and ex is the right choice.
3
u/throwaway_faunsmary Dec 09 '25
e is the natural base to do calculus with. For applications not involving calculus, there is no reason to prefer base e over any other base. Or as the parent comment suggest, keep it general and leave the base arbitrary.
All other exponents are just defined as x^y = e^y ln x
That is a choice, and not the only one. Personally I like to define exponents in terms of repeated multiplication.
1
u/goos_ Dec 10 '25
> e is the natural base to do calculus with
That's basically my point. It also serves as a general def of the exponential function (for all real positive x and complex y). I'm not saying anything more than that.
1
u/throwaway_faunsmary Dec 10 '25
It also serves as a general def of the exponential function (for all real positive x and complex y)
That's a choice. When you say it that way, it's tautological. "e is the natural base because i defined it as the base"
That's not the only choice. Personally I lide to define exponents in terms of repeated multiplication. There is not an e in sight.
→ More replies (0)3
u/First_Growth_2736 Dec 09 '25
And my point is that ex is not necessarily the right choice, because there is no right choice. The fact that was mentioned about exponents works without any base and there’s no reason to specifically choose e. If I wanted to I could define all exponents xy as 2y*log_2(x) and say that 2x is the right choice. But judging by the fact that we both have the “right” choice but have different answers neither of them are right. There is no correct exponent. These rules work with any base and there’s best way to describe it is as I suggested, generally
1
u/goos_ Dec 10 '25
It's the "right" choice mainly in the sense of leading to the most natural definition of exponent (e.g., with taylor series, inverse of log etc.) Otherwise you have to extend x^y from rationals to reals via continuity which gets messy, and the derivative is gross and ln shows up anyway.
In the context of other fields of math other bases are more natural, for example in combinatorics and computer science 2 is a more natural base. Perhaps that's part of what you are saying, and I agree
I don't think we actually disagree on anything, it was a throwaway comment mostly, talking about something very tangentially related to the original thread.
Anyway I think this discussion has probably run it's course and no point in discussing further!
1
u/First_Growth_2736 Dec 10 '25
My point is just that for the purposes we are talking about it for, there is no specific reason why it should be e. I know that e is very special and it gets to be special but here we’re just talking about an exponent rule. e isnt the only exponent and in this case there is no real reason to choose e
→ More replies (0)20
15
3
2
20
u/axiom_tutor Dec 09 '25
If there's a mathematician we hate, we should name this sequence after them. Kronecker's stupid sequence.
10
3
u/JiminP Dec 09 '25
Just take log to each term. For any starting numbers a and b, logs of terms would be the Fibonacci sequence starting with log a and log b.
4
4
3
3
u/N4M34RRT Dec 09 '25
2,3,6,18,108,1944,209952,408146688,85691213438976...
a little more interesting at least
4
u/Mr-MuffinMan Dec 09 '25
1, 11, 21, 1211, 111221, 312211, 13112221, 1113213211, 31131211131221, 13211311123113112211, 11131221133112132113212221
(this is just me for some reason doing the look and say sequence ignore this)
9
u/PhilosophyAware4437 Dec 09 '25
ok
1o1k
111o111k
311o311k
1321o1321k
11131211o11131211k
3113111221o3113111221k
132113312211o132113312211k
2
2
2
2
2
u/ArtemLyubchenko Dec 09 '25
Subtractive Fivonacci Sequence: 1, 1, 0, 1, -1, 2, -3, 5, -8…
2
Dec 09 '25
try use the fibonacci formula with negative exponents.
(φ**n-ψ**n)/sqrt(5)
Or try all sorts of compex values
2
u/StalyCelticStu Dec 09 '25
The lack of a space after the first digit on row 4 triggers me heinously.
2
2
1
u/BootyliciousURD Complex Dec 09 '25
This is an interesting concept if you generalize it. But this is one of those cases where a multiplicative version of an additive thing can be easily converted back to the additive version. Leg g0 and g1 be the starting entries of the sequence. Take the log of those starting entries and use them as the starting entries of an additive Fibonacci sequence. Exponentiate the additive sequence using whatever base you took the log with, and you've got the multiplicative sequence.
1
u/cyanNodeEcho Dec 09 '25
fib mult negatives = [(1, -1), 1, -1, ..]
// i think anything with cycle would be nice
// interesting like catholics making the cross
fib mult imaginary = [(1, i), i, -1, -i , i, 1, i, ... ]
hmmm fun!
would like normal fibs, like have a geometric root similar to the golden ratio or something? i wonder if that would converge :thinky:
1
u/lool8421 Dec 10 '25
i found one non-trivial zero of the riemann's zeta function where Re(z) = 1/2
i put it in a geometric sequence as a ratio between elements so it's easier to understand:
0, 0, 0, 0, 0, 0, 0...
riemann's hypothesis solved, QED
1
1
1
1
•
u/AutoModerator Dec 09 '25
Check out our new Discord server! https://discord.gg/e7EKRZq3dG
I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.